Tag: geometry

Position is easy, just represent it with a point in 3D space. But how do you specify its orientation — which direction it’s pointing?

At first glance, it seems a vector will do. After all, a vector points in some direction, right? If the plane is pointing east, represent its orientation by a unit vector pointing east.

Unfortunately, we quickly run into trouble when we try to roll. If we’re facing east, and we roll 90 degrees, we’re still facing east. Clearly we’re missing something.

Euler Angles

When real pilots talk about their orientation, they talk about roll, yaw, pitch. Pitch is going up or down, yaw is going left or right, roll is, well, roll.

Any change in orientation can be described by some combination of roll, yaw, pitch. This is the basis for Euler Angles. We use three angles to represent the airplane’s orientation.

This is all fine and dandy if we want to represent the orientation of a static object in space. But when we try to adjust our orientation, we start to run into problems.

You’re thinking, this should be simple! When we turn left or right, we just increment the yaw variable, right? Yes, it seems to work, at least initially. You can turn left and right, up and down, and roll around.

Implement it in Unity and play around a bit, however, and you begin to notice that things don’t quite behave the way you expect.

In this animation, I’m holding down the right button:

The plane does rotate to the right, but it’s not rotating relative to itself. Instead it’s rotating around some invisible y-axis. If it was rotating relative to itself, the green arrow shouldn’t be moving.

The problem becomes more and more severe when the pitch of the plane becomes higher and higher. The worst case is when the airplane is pointing straight up: then roll and yaw become the same thing! This is called gimbal lock: we have lost a degree of freedom and we can only rotate in 2 dimensions! Definitely not something desirable if we’re controlling a plane or spaceship.

It turns out that no matter what we do, we will suffer from some form of gimbal lock. As long as we use Euler Angles, there is one direction where if we turn too far, everything starts to screw up.

Practical Introduction to Quaternions

All is not lost, however. There is a way to represent orientation that represents all axes equally and does not suffer from gimbal lock. This mythical structure is called the quaternion. Unlike Euler Angles which describe your orientation relative to a fixed set of axes, quaternions do not rely on any fixed axis.

The drawback is that quaternions are unintuitive to understand for humans. There is no way to “look” at a quaternion and be able to visualize what rotation it represents. Fortunately for us, it’s not that difficult to make use of quaternions, even if we can’t visualize quaternions.

There is a lot of theory behind how quaternions work, but in this article, I will gloss over the theory and give a quick primer to quaternions, just the most common facts you need to use them. At the same time, I will implement the operations I describe in C#, so I can integrate them with Unity. If you don’t know C#, you can freely ignore the code.

Definition

A quaternion is an ordered pair of 4 real numbers (w,x,y,z). We write this as

The letters i,j,k are not variables. Rather, they are independent axes. If you like, you can think of the quaternions as a 4 dimensional vector space.

The defining property of quaternions is:

Play around with it a bit and you can derive 6 more identites:

If you’ve worked with complex numbers, this should seem familiar. Instead of 2 parts of a complex number (the real and imaginary parts), we have 4 parts for a quaternion.

The similarity doesn’t end here. Multiplying complex numbers represents a rotation in 2 dimensions. Similarly, multiplying by a quaternion represents a rotation in 3D.

One curious thing to note: we have and . We switched around the terms and the product changed. This means that multiplying quaternions is kind of like multiplying matrices — the order matters. So multiplication is not commutative.

In Unity, the input is not given to us as a single true/false value, but a float between -1 and 1. So holding right increases the LeftRight input gradually until it reaches 1, avoiding a sudden jump in movement.

What’s ToUnityQuaternion? Well, it turns out that Unity already has a Quaternion class that does everything here and much more, so all this could have literally been implemented in one line if we wanted.

Anyways, let’s see the result.

As you can see, holding right turns the plane relative to itself now, and the green arrow stays still. Hooray!

We have a square, and we put a point on each side of the square. Then we connect the four points to create a quadrilateral. How can we make this quadrilateral have the smallest possible perimeter?

Intuitively, you may believe that this natural, obvious configuration should produce the least perimeter:

Attempt with Calculus

How can we prove that this indeed gives us the smallest possible perimeter?

A first attempt might be to give variables to the side lengths, and somehow find the minimum perimeter using algebra and calculus tools. So there are four independent points — let’s parameterize them with four variables, and assume the side length of the square is 1:

Then we want to minimize this expression:

At this point, it isn’t clear how to proceed — there doesn’t seem to be any way to minimize this expression of four variables.

Proof by Net

We’ll have to try something different. It’s hard to make sense of anything when there are four independent variables. Instead, if we expand things out a bit, things start to become more manageable:

What we did was reflect the square three times, and each time the square is reflected, the inscribed quadrilateral goes with it. By taking only the relevant parts of the quadrilateral, we get the green path.

Now we might have a solution. If we had a different green path, can we reverse the steps and get the original quadrilateral back? Basically, the following requirements have to be met:

The path has to cross all three of the internal lines BC, BA, and DA.

The path’s position on the bottom-most line, DC must be the same when reflected onto the top-most line DC.

With these requirements in mind, the shortest green path that satisfies these requirements is a straight line connecting a point on the bottom left to its reflected point on the top right:

Our intuition at the start was well-founded.

Now notice that this isn’t the only possible shortest path. If we move the entire green line to the left or right, we get a different path of the same length!

For instance, the degenerate ‘quadrilateral’ formed by connecting two opposite corners has the same perimeter as the one we get by connecting the midpoints. Neat, huh?

For many people (for me at least), the Harmonic Conjugate is a difficult concept to understand. I didn’t really get it the first time I saw it, at Mathcamp. Let’s take the definition of the harmonic conjugate:

AB and CD are harmonic conjugates if this equation holds:

If you’re like me, you’re thinking along the lines of “But why? Why is this defined this way? Why would we spend so much time proving things about this weird concept? What’s the point, what’s the use?”

Even now, I can’t really give you an intuitive explanation of why this equality is so important. On the other hand, I could certainly come up with a few problems in which the concept of the harmonic conjugate turns to be useful.

Apollonius and Fleeing Ships

Apollonius’s problem was this: you are in control of a ship (point A on diagram), and you are in pursuit of another ship (point B). The other ship is fleeing in a straight line in some direction:

Your speed is (obviously) faster than the speed of the other ship: say they’re going at 30 km/h and you’re going at 50 km/h. Additionally, your ship is required to go in a straight line.

In which direction should you set off in order to intercept the fleeing ship?

Solution with Harmonic Conjugates

The first step of the solution is to construct harmonic conjugates CD so that their ratio is the ratio of your speed to the other ship’s speed (we’ll prove later that this is actually possible; assume we can do this for now):

Next, draw a circle with diameter CD:

There is a point where the ray from B (their ship) intersects this circle. Now go to this point immediately, in a straight line: the ship will be there.

The Proof

In order to prove that this works, we’ll need to take a step back and look at how we constructed the points C and D. The solution turns out to be evident directly from the construction of the harmonic conjugates.

Again, let’s assume our desired ratio is 5/3. Starting with the points A and B, the first step is constructing some point P so that:

This is fairly easy to do. Draw a circle of radius 5 around A, and draw a circle of radius 3 around B — the intersection P of these two circles forms the correct ratio. (if the circles don’t intersect, just scale everything up and try again)

Next, dropping the internal and external angle bisectors of the new triangle gives the harmonic conjugates C and D:

The angle between any two edges of the tetrahedron is , so it’s easy to (falsely) conclude that the angle between two faces must also be .

But this isn’t the case: the plane angle is defined as the angle between two lines on the two planes that are both perpendicular to the edge. None of the edges in a tetrahedron is perpendicular to any other, so the answer of is invalid.

We can try to compute the angle using regular Euclidean solid geometry, but things tend to get messy. A different way to approach the problem is by using vector geometry: using vector methods we can easily calculate the plane angle of the tetrahedron (as well as the icosahedron and the dodecahedron).

Assume symmetry. We represent three concurrent edges of the polyhedron as three vectors beginning at the same point: , , and ; let and be the angles between the vectors (by symmetry we’re assuming that the two alpha’s are equal):(in case my poor drawing skills does not make this apparent, vectors a and b form one face of the polyhedron and c and b form another face)

For simplicity, let’s also say the lengths of each of the three vectors is 1.

We want to compute the angle between the plane formed by and , and the plane formed by and . Hence let and be the perpendiculars to and respectively each ending at the same point as their respective vectors:

For any two vectors, the dot product is defined with being the angle between the vectors. Given that the lengths of the vectors are all 1, we have:

Also, by vector addition:

Hence and .

We want to find the angle between and — call this angle . Then

The dot product of vectors x and y is simply:

Additionally . Hence the cosine of the angle is:

We can now use this newly derived formula to calculate plane angles! For example…

Tetrahedron

In the tetrahedron, both and are 60:So and .

Icosahedron

In the icosahedron, but :The top ‘cap’ is a regular pentagon, which has a vertex angle of 108; each side of the pentagon constitutes a side of an equilateral triangle. Since , , and .

Dodecahedron

Computing angles for the dodecahedron works a bit differently from the tetrahedron and icosahedron. Instead of using existing edges as vectors, we construct an equilateral triangle by connecting three vertices:

Today I was reading some math book when the author mentions Varignon’s theorem, and gives a proof. The proof was not very long, but it was somewhat confusing. On Wikipedia, several more short proofs were given, but they were all more confusing than need be.

I remembered seeing the theorem proven using vector geometry before, but I couldn’t find the text (nor any other page / book that proves it this way) —

[image shamelessly taken from Wikipedia]

Varignon’s theorem states that in any quadrilateral, if we join the midpoints of the sides, then we get a parallelogram.

In the diagram, it suffices to prove that vector HG is equal to vector EF — vectors must have both the same orientation and length to be equal. This works since any method that proves HG = EF can also prove HE = GF. The proof goes as follows —

I wrote the Euclid contest this year two days ago, on tuesday. There were 10 problems, and the tenth problem was a nice geometry problem. Three subproblems with a nice neat triangle on the right, with the subproblems getting progressively harder and harder. As I proceeded with the problem, I couldn’t help but imagine it as a Project Euler problem — instead of finding one such integer triangle, it would ask you to find the sum of the perimeters of all such integer triangles with perimeter less than 10^12 or some large number.

A modified problem

In the above diagram, and . Let , , , , and . Write an algorithm to find triangles satisfying these conditions where a, b, c are all integers.

Similar triangles

It is difficult to try to find integer triangles with such strange requirements as these. It seems that the line is completely unnecessary, but if we take it out, there doesn’t seem to be any way to relate the angle ratios to integer side lengths.

We can prove that is similar to . Being an exterior angle, , and also . Both of the triangles have an angle of measure and another angle of measure , thus they are similar.

From the similar triangles ratio

We can write d in terms of the three sides of the triangle:

Similarly, the side can be written as . Then we have the ratio

Solving for x allows us to express it in terms of the three sides of the triangle, again:

Constructing another similar triangle

Our goal here is to relate the lengths a, b, c with a simple equation, which then the problem turns into a number theory problem. Since we can write the lengths d and x in terms of a, b, and c, we can also relate any of a, b, c, d, x in an equation.

Again, there doesn’t seem to be a way to relate all of the variables together, in a way that any solution implies the original angle ratio required — unless we make a construction.

Here we extend AB to F, so that and is an isosceles triangle.

Again, since the exterior angle here is , both . Also with this construction, , and is also isosceles, hence .

With this construction, we can write the ratio

Perfect! Cross multiplying and then substituting in the original sides of the triangles gives

Simplifying this gives

Number theory magic

Now that we have an equation relating the three side lengths — it’s easy to verify that any three integers satisfying the triangle inequality gives a triangle with the required conditions — we can use number theory to try to generate integers that fit the equation.

If we expand the equation, we get

It makes sense to solve for c, which can be done using just the quadratic formula:

We need the discriminant D — the expression inside the square root — to be a perfect square, where we are allowed to have integer values for a and b. If we can get D to be a perfect square, then c will turn out to be a rational number. Then multiplying all three variables by a constant gives integer values for all three.

So we defined D:

Expanding this gives

Fortunately, this expression has an interesting factorization:

Or we can also write

We’ve simplified this problem to finding values where is a perfect square, that is:

This is just finding Pythagorean triples where one of the two sides are even! For instance, in the triple (3,4,5), we have a=3 and b=2. However, substituting a=3, b=2 into the quadratic formula gives c=5. This is almost a solution, only that the sides have to satisfy the triangle inequality (two sides have to add up to more than the third side).

The next Pythagorean triple (5,12,13) gives a=5 and b=6. Substituting this in gives c=11/9, which does satisfy the triangle inequality. Multiplying everything by 9 gives a=45, b=54, c=11 as the smallest working combination.

With this method, it is possible to quickly find arbitrarily many such triples, using Pythagorean triples as a starting point (which can be generated quickly with known methods).

The equation of a hyperbola gives it some very interesting properties under a series of stretches. Consider the hyperbola :

If we stretch the hyperbola horizontally by a factor of 2 about the x axis — replacing in the equation with , we have the equation .

Now, if we compress the resulting hyperbola vertically by a factor of 2 — replacing in the equation with and simplifying, we get the original hyperbola, . So by a horizontal stretch followed by a vertical compression, we get the original hyperbola.

However, consider a point on the hyperbola , say . Stretching the hyperbola horizontally by a factor of 2, this point gets transformed to . Even though each point has a different image point, however, the graph as a whole remains identical.

More generally, if we stretch by a factor of instead of 2, where is any positive number, we still obtain the original hyperbola after the two stretches. The point gets transformed to the point . We call this combinations of transformations a hyperbolic rotation. Intuitively, the points on the hyperbola are ‘rotated’ downwards.

There are some interesting things about the hyperbolic rotation. For one, by choosing the right for the transformation, it is possible to transform any point on the hyperbola into any other point on the same arm of the hyperbola (by allowing to be smaller than 1). Also, since the hyperbolic rotation is composed of two simple stretches, parallel lines as well as ratios of a line segments are preserved. The area of a figure is also preserved since one stretch reduces the area, while the other multiplies the area by the same amount.

With these facts, we can prove things about the hyperbola itself:

Theorem: Let P be a point on the hyperbola . Let be the line passing through P tangent to the hyperbola. Let X be the intersection of and the x-axis, and Y be the intersection of with the y-axis. Prove that P is the midpoint of XY.

The proof is very simple with the hyperbolic rotation. Consider when . The theorem obviously holds here, because of symmetry. But by applying a hyperbolic rotation, we can transform the point into any other point on the hyperbola. Since ratios between line segments don’t change with a hyperbolic rotation, P is always the midpoint, and we are done.